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(B) Given $a 	imes b = 2a 	imes c$,we can write $a 	imes (b - 2c) = 0$. This implies that the vector $(b - 2c)$ is parallel to $a$. Given $b - 2c =

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$\pm 4$

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(B) Given $a 	imes b = 2a 	imes c$,we can write $a 	imes (b - 2c) = 0$. This implies that the vector $(b - 2c)$ is parallel to $a$. Given $b - 2c = \lambda a$,we have $|b - 2c|^2 = \lambda^2 |a|^2$. Expanding the left side: $|b|^2 + 4|c|^2 - 4(b \cdot c) = \lambda^2 (1)^2$. Given $|b| = 4$ and $|c| = 1$,we have $16 + 4(1) - 4(b \cdot c) = \lambda^2$,so $20 - 4(b \cdot c) = \lambda^2$. We are given $|b 	imes c| = \sqrt{15}$. Since $|b 	imes c| = |b||c| \sin \alpha = \sqrt{15}$,where $\alpha$ is the angle between $b$ and $c$,we have $(4)(1) \sin \alpha = \sqrt{15}$,so $\sin \alpha = \frac{\sqrt{15}}{4}$. Then $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{15}{16} = \frac{1}{16}$,so $\cos \alpha = \pm \frac{1}{4}$. Since $b \cdot c = |b||c| \cos \alpha = (4)(1) \cos \alpha = 4 \cos \alpha$,we have $b \cdot c = 4(\pm \frac{1}{4}) = \pm 1$. Substituting this into the equation for $\lambda^2$: $\lambda^2 = 20 - 4(\pm 1) = 20 \mp 4$. If $b \cdot c = 1$,$\lambda^2 = 16$,so $\lambda = \pm 4$. If $b \cdot c = -1$,$\lambda^2 = 24$,which is not among the options. Thus,$\lambda = \pm 4$.

Let $a, b, c$ be three vectors such that $a \neq 0$,$a \times b = 2a \times c$,$|a| = |c| = 1$,$|b| = 4$,and $|b \times c| = \sqrt{15}$. If $b - 2c = \lambda a$,then $\lambda$ equals:

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